search for: iris

...invalid permissions" as part of a segfault).? Yes, I realize this is a silly example that you wouldn't actually do, but... there may be other more interesting cases with the same problem.? The following was R-2.10.0 on Linux (with a Windows-64 2.11.0 difference to follow below): > data(iris) > split(1:nrow(iris), list(iris[,1], iris[,2])) ?*** caught segfault *** address 0x7fc806cd3d0c, cause 'invalid permissions' Traceback: ?1: split.default(1:nrow(iris), list(iris[, 1], iris[, 2])) ?2: split(1:nrow(iris), list(iris[, 1], iris[, 2])) Possible actions: 1: abort (with cor...

It's in the Venables & Ripley MASS (ed 3) book in the section on principal components. The context is as follows > ir <- rbind(iris[,,1], iris[,,2], iris[,,3]) > ir.species <- factor(c(rep("s",50),rep("c",50),rep("v",50))) (then they use brush(ir) which I guess is not an R function) and then > princomp(log(ir[1:4]),cor=T) (there is no [1:4] in the text which is probably a typo) >From...

I have a dataset and I wish to use two different models to predict. Both models are SVM. The reason for two different models is based on the sex of the observation. I wish to be able to make predictions and have the results be in the same order as my original dataset. To illustrate I will use iris: # Take Iris and create a dataframe of just two Species, setosa and versicolor, shuffle them data(iris) iris <- iris[(iris$Species=="setosa" | iris$Species=="versicolor"),] irisindex <- sample(1:nrow(iris), nrow(iris)) iris <- iris[irisindex,] # Make predictions on se...

...d, which swaps one of the plot() calls for hist(), results in a 4 page pdf with one plot per page. How might I get the histogram with 3 other scatter plots onto a single pdf page? platform powerpc-apple-darwin8.10.1 version.string R version 2.6.1 (2007-11-26) Thanks! Ben ###BEGIN data(iris) orig.par = par(no.readonly = TRUE) pdf(file = "just_plots.pdf") par(mfrow=c(2,2)) plot(iris$Sepal.Length, iris$Sepal.Width, main = "Plot 1") plot(iris$Petal.Length, iris$Petal.Width, main = "Plot 2") plot(iris$Sepal.Length, iris$Petal.Length, main = "Plot 3&q...

The following command produces red axis line in a pairs plot: pairs(iris[1:4], main = "Anderson's Iris Data -- 3 species", pch = "+", col = c("red", "green3", "blue")[unclass(iris$Species)]) Trying to fool pairs in the following way produces the same plot as above: pairs(iris[1:4], main = "Anderson's...

...ions in a loop and then plotting the results by adding a point to each of 2 charts at the end of the loop. Its very informative as you can see the progression through time. My problem is, if I have 2 plots, I don't know how to get the focus back to the first plot. layout(matrix(c(1,2))) plot(iris[,1],col="red",) #plot1 plot(iris[,3],col="blue") #plot2 #goes on plot2 lines(iris[,2],col="pink") #how do I put this line on plot 1 lines(iris[,4],col="black") I tried the method below but when you switch the focus back to screen 1 the line gets drawn n...

Hi folk, I tried this and it works just perfectly tapply(iris[,1],iris[5],mean) but, how to obtain a single table from multiple variables? In tapply x is an atomic object so this code doesn't work tapply(iris[,1:4],iris[5],mean) Thanx and great summer holidays Gianandrea -- View this message in context: http://www.nabble.com/multiple-tapply-tp1886806...

Hi R People: Several of you pointed out that using "tapply" on a data frame will work on the iris data frame. I'm still having a problem. The iris data frame has 150 rows, 5 variables. The first 4 are numeric, while the last is a factor, which has the Species names. I can use tapply for 1 variable at a time: >tapply(iris[,1],iris[,5],mean) setosa versicolor virginica 5.006...

Dear all, I'd like to get a percentage variable based on a group, but without creating a new data frame. For example: data(iris) iris$percent <-unlist(tapply(iris$Sepal.Length,iris$Species,function(x) x/sum(x, na.rm=TRUE))) This does not work, I should have only three standard values, respectively for setosa, versicolor, and virginica. How can I do this? MANY THANKS, Karine [[alternative HTML version d...

Hi Dieter and R community: I tried both of these three versions with ylim as suggested, none work: I am getting only single (pch = 16) not overlayed (pch =3) everytime. *vs 1* require(lattice) xyplot(Sepal.Length ~ Sepal.Width | Species , data= iris, panel= function(x, y, subscripts) { panel.xyplot(x, y, pch=16, col = "green4", ylim = c(0, 10)) panel.lmline(x, y, lty=4, col = "green4") panel.xyplot (iris$Petal.Length [subscripts], iris$Petal.Width[subscripts], pch= 3, col = "red") panel.lmline(iris$Petal.Le...

Hopefully I will not be flamed for this on the list, but I am starting out with R and having some trouble with combining plots. I am playing with the famous iris dataset (checking out example dataset in R while reading through Introduction to datamining) What I would like to do is create three graphs (combined boxplots) besides each other for each of the three species (Setosa, Versicolour and Virginica) with each graph showing showing four boxplots Sepal.L...

Por si alguno pudiera ayudarme. Al realizar el t.test para una muestra, junto con el valor de t y el p-valor, la función proporciona la estimación de la media y su INTERVALO DE CONFIANZA. Desde el punto de vista de la estadística de rangos esto se puede hacer mediante: > iris$MEDIANA <- with(iris, 2.95) > median(iris$Sepal.Width - iris$MEDIANA, na.rm=TRUE) # median difference [1] 0.05 > wilcox.test(iris$Sepal.Width, iris$MEDIANA, alternative=''two.sided'',paired=TRUE) Wilcoxon signed rank test with continuity correction data: iris$Se...

Compiled great, got both my RSA and DSA keys' generated for Protocol 1/2, started up fine ... try to connect and get a bunch of errors: May 11 14:01:47 iris sshd[8578]: error: Couldn't wait for child '/bin/ls -alni' completion: No child processes May 11 14:01:47 iris last message repeated 3 times May 11 14:01:47 iris sshd[8578]: error: Command '/bin/ls -alni': select() failed: Interrupted system call May 11 14:01:47 iris sshd[8578]:...

Hi, I am a newbie of R, and playing with the "ifelse" statement. I have the following codes: ## first, for(i in 1:3) { for(j in 2:4) { cor.temp <- cor(iris.allnum[,i], iris.allnum[,j]) if(i==1 & j==2) corr.iris <- cor.temp else corr.iris <- c(corr.iris, cor.temp) } } this code is working fine. I also tried to perform the same thing in another way with "ifelse": for(i in 1:3) { for(j in 2:4) { cor.temp <- cor(iris.allnum[,i], i...

Greetings, I have been exploring the use of the caret package to conduct some plsda modeling. Previously, I have come across methods that result in a R2 and Q2 for the model. Using the 'iris' data set, I wanted to see if I could accomplish this with the caret package. I use the following code: library(caret) data(iris) #needed to convert to numeric in order to do regression #I don't fully understand this but if I left as a factor I would get an error following the summary fu...

Hi there: I have a question about generating mean value of a data.frame. Take iris data for example, if I have a data.frame looking like the following: --------------------- Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0...

Hi, I tried to use naivebayes in package 'e1071'. when I use following parameter, only one predictor, there is an error. > m<- naiveBayes(iris[,1], iris[,5]) > table(predict(m, iris[,1]), iris[,5]) Error in log(sapply(attribs, function(v) { : Non-numeric argument to mathematical function However, when I use two predictors, there is not error any more. > m<- naiveBayes(iris[,1:2], iris[,5]) > table(predict(m, iris[,1:2...

Hello all, I have a large data frame and I want to look at the distribution of each variable very quickly by plotting an individual histogram for each variable. I'd like to do so using lattice. Here is a small example using the iris data set: histogram(as.formula(paste("~",paste(colnames(iris[,!sapply(iris,is.factor)]),collapse="+"))),data=iris[,!sapply(iris,is.factor)]) However in this graphic, the x-axis and the breaks are the same for all 4 variables. I would like them to be adjusted to the data in eac...

Just out of curiosity, I took the default "iris" example in the RF helpfile... but seeing the admonition against using the formula interface for large data sets, I wanted to play around a bit to see how the various options affected the output. Found something interesting I couldn't find documentation for... Just like the example... &gt...

..., the AUC of the training set is coming to be 1. Is this always the case with random forests? Can someone please clarify this? I have given a simple example, first using logistic regression and then using random forests to explain the problem. AUC of the random forest is coming out to be 1. data(iris) iris <- iris[(iris$Species != "setosa"),] iris$Species <- factor(iris$Species) fit <- glm(Species~.,iris,family=binomial) train.predict <- predict(fit,newdata = iris,type="response") library(ROCR) plot(performance(prediction(train.predict,iris$Species),&qu...